# Inequality

• November 15th 2007, 07:11 AM
james_bond
Inequality
Prove that if $ab+bc+ca=1$ than $\frac{1-a^2}{1+a^2}+\frac{1-b^2}{1+b^2}+\frac{1-c^2}{1+c^2}\le\frac{3}{2}$. ( $a,b,c\in\mathbb{R}$)
• November 16th 2007, 11:28 AM
Opalg
Quote:

Originally Posted by james_bond
Prove that if $ab+bc+ca=1$ than $\frac{1-a^2}{1+a^2}+\frac{1-b^2}{1+b^2}+\frac{1-c^2}{1+c^2}\le\frac{3}{2}$. ( $a,b,c\in\mathbb{R}$)

Who thinks these horrible things up (and why do I spend hours brooding over them)?

I'll assume that a, b and c are all positive (the problem's hard enough as it is, and I don't want to be bothered with negative quantities).

Notice first that $\frac{1-a^2}{1+a^2} = \frac{2-(1+a^2)}{1+a^2} = \frac2{1+a^2}-1$ (and similarly for b and c), so the inequality can be written

. . . . . $\frac1{1+a^2}+\frac1{1+b^2}+\frac1{1+c^2} \leqslant\frac94$ . . .(1)

If bc+ca+ab=1 then

. . . . . $1+a^2 = (bc+ca+ab)+a^2 = (a+b)(a+c)$ . . .(2)

(with similar inequalities for b and c).

Also, $a+b+c = (a+b+c)(bc+ca+ab) = (b+c)(c+a)(a+b) + abc$ . . .(3)

(just multiply out both sides to verify that).

By the arithmetic-geometric mean inequality,

. . . . . $abc = \sqrt{bc}\sqrt{ca}\sqrt{ab} \leqslant \frac{b+c}2\frac{c+a}2\frac{a+b}2 = {\textstyle\frac18}(b+c)(c+a)(a+b)$ . . .(4)

(that's the only place where I assume that a, b and c are positive).

It follows that

. . . . . $\frac1{1+a^2}+\frac1{1+b^2}+\frac1{1+c^2} = \frac1{(a+b)(a+c)} + \frac1{(b+c)(b+a)} + \frac1{(c+a)(c+b)}$ . . .by (2)

. . . . . . . . . . . . . . . . . . . . . . . $= \frac{2(a+b+c)} {(b+c)(c+a)(a+b)}$

. . . . . . . . . . . . . . . . . . . . . . . $= \frac{2(b+c)(c+a)(a+b) + 2abc}{(b+c)(c+a)(a+b)}$ . . .by (3)

. . . . . . . . . . . . . . . . . . . . . . . $= 2 + \frac{2abc}{(b+c)(c+a)(a+b)}$

. . . . . . . . . . . . . . . . . . . . . . . $\leqslant 2 + \textstyle\frac14 = \frac94$ . . .by (4).

By (1), that was what we wanted to prove.